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MATH 117 AppliedStatistics Monday, 3512 Project 3 is due TODAY; online assignment 10.1 is due Tuesday, March 6; and assignment 11.2 is due Wednesday, March 21. To complete an hypothesis test: 1. Determine H o and H a, 2. Find phat, 3. Find the standard error, 4. Find the zscore, 5. Find the Pvalue (from the table), 6. Compare to the level of significance (reject if Pvalue < level of sig. fail to reject if Pvalue > level of sig.) Title: Aug 28 6:06 AM (1 of 20)

MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

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Page 1: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

MATH 117 ­ Applied StatisticsMonday, 3­5­12

Project 3 is due TODAY; online assignment 10.1 is due Tuesday, March 6; and assignment 11.2 is due Wednesday, March 21.

To complete an hypothesis test:

1. Determine Ho and Ha,

2. Find p­hat,

3. Find the standard error,

4. Find the z­score,

5. Find the P­value (from the table),

6. Compare to the level of significance (reject if P­value < level of sig. fail to reject if P­value > level of sig.)

Title: Aug 28 ­ 6:06 AM (1 of 20)

Page 2: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Title: Feb 21 ­ 8:25 PM (2 of 20)

Page 3: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Title: Oct 3 ­ 4:12 PM (3 of 20)

Page 4: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

1. A random sample of 70 homes in a community revealed that 23 of the homes were rentals. Find a 90% confidence interval for the proportion of all homes in the community that are rentals. Keep at least 3 significant digits.

a. Point estimate = 23/70 = .32857

b. SE = SQR(.32857(1‐.32857)/70) = .056139

c. z‐score (from table) = 1.645

d. ME = 1.645*.056139 = .0923

e. Confidence interval: .236 < p < .421

Title: Mar 4 ­ 9:17 AM (4 of 20)

Page 5: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

2. A random sample of 600 households discovered that 473 of them had tuned into the Super Bowl. Find a 95% confidence interval for the proportion of all households that had watched the Super Bowl.

a. Point estimate = 473/600 = .78833

b. SE = SQR(.78833(1‐.78833)/600) = .016677

c. z‐score (from table) = 1.96

d. ME = 1.96*.016677 = .03269

e. Confidence interval: .7556 < p < .8210

f. If there are approximately 126,000,000 households in the US, find the interval estimate for the total number of households tuned in to the Super Bowl.

95,205,600 < # of households < 103,446,000

Title: Mar 4 ­ 9:17 AM (5 of 20)

Page 6: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

3. In the 1992 presidential election an estimated 68% of the minority voters voted Democratic. In an exit poll of 285 minority voters in the 2008 election, 204 voted Democratic. Is this sufficient evidence at the 5% significance level to conclude that the percentage has increased?

a. Ho: p = .68

Ha: p > .68

b. p‐hat = 204/285 = .716

c. SE = SQR( .68(1 ‐ .68)/285) = .0237

d. z‐score = (.716 ‐ .68)/.0237

e. P‐value (from table) = .0968 > .05

f. Conclusion: Do not reject Ho. Maybe the percentage really is .68.

Title: Feb 24 ­ 8:41 PM (6 of 20)

Page 7: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

4. The electronics industry claims that 18% of all TV's now sold are 3‐D capable. In a poll of 9 area stores we discovered that last week there were 41 3‐D models and 239 non‐3‐D models sold. Is this sufficient evidence at a 10% level of significance to conclude that the percentage of 3‐D models is actually less than the 18% claimed.

a. Ho: p = .18

Ha: p < .18

b. p‐hat = 41/280 = .146

c. SE = SQR(.18(1 ‐ .18)/280) = .0230

d. z‐score = (.146 ‐ .18)/.0230 = ‐1.48

e. P‐value (from table) = .0694 < .10

f. Conclusion: Reject Ho. We have enough evidence to conclude that p < .18.

Title: Feb 24 ­ 8:41 PM (7 of 20)

Page 8: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

5. A Chinese shoe manufacturer has guaranteed Nike that they will produce Nike "Free Run" shoes with no more than 6% defective shoes. Note that the defective shoes will still be sold as seconds at the outlet stores. Nike tests a sample of 850 pairs of shoes and finds 58 pairs are defective. The manufacturer claims that this was simply an exceptionally bad sample and the percentage of all shoes is still 6%. Is there sufficient evidence at the 10% level of significance to refute the 6% claim?a. Ho: p = .06

Ha: p > .06

b. p‐hat = 58/850 = .0682

c. SE = SQR(.06(1 ‐ .06)/850) = .00815

d. z‐score = (.0682 ‐ .06)/.00815 = 1.01

e. P‐value (from table) = .1562 > .10

f. Conclusion: Do not reject. There is insufficient evidence to refute the .06 claim.

Title: Feb 26 ­ 12:55 PM (8 of 20)

Page 9: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

6. A trivia game claimed that 20% of M&M's in a typical assortment are red. We set out to refute that claim by showing that the proportion is either greater than or less than 20%. We buy a case of M&M's and find that 118 of the 685 M&M's are red. Is this enough evidence to refute the 20% claim at a .05 level of significance? Note that this involves a two‐tailed P‐value.

a. Ho: p = .20

Ha: p is not equal to .20

b. p‐hat = 118/685 = .172

c. SE = SQR(.2(1 ‐ .2)/685) = .0153

d. z‐score = (.172 ‐ .20)/.0153 = ‐1.83

e. P‐value (from table) = .0336 x 2 tails = .0672 > .05

f. Conclusion: Do not reject. There is insufficient evidence to refute the 20% claim.

Title: Feb 26 ­ 12:56 PM (9 of 20)

Page 10: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Title: Mar 4 ­ 9:19 AM (10 of 20)

Page 11: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Title: Mar 4 ­ 9:17 AM (11 of 20)

Page 12: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Suppose a poll was taken as people left the theater after seeing "The Artist". 45 out of 62 women gave the movie a 4 out of 4 rating and 37 of 58 men gave it the same rating. What is the P­value for this 2­proportion z­test? Is this sufficient evidence at the 10% level of significance that women like the movie better than men?

Title: Mar 4 ­ 9:20 AM (12 of 20)

Page 13: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Now we look at two categorical variables and try to determine if one is dependent on the other. First, identify the potential explanatory variable. Then construct a contingency table and calculate the conditional percentages and examine the conditional distributions.

Title: Mar 4 ­ 9:20 AM (13 of 20)

Page 14: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Recall the contingency table used to compare the categorical variable of income with the categorical variable of happiness.

It's much easier to compare if we calculate the percentages.

Title: Mar 4 ­ 9:21 AM (14 of 20)

Page 15: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Do the distributions indicate that happiness is dependent on income?

Title: Mar 4 ­ 9:21 AM (15 of 20)

Page 16: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Consider the conditional percentages for gender as the explanatory variable and happiness as the response variable. Admittedly these numbers are "fudged" for demonstration.

Title: Mar 4 ­ 9:22 AM (16 of 20)

Page 17: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

If the conditional distributions are not identical but are very similar, we suspect that the difference is due to the "luck" of the sample rather than to a real dependence. But how do we determine if the difference is significant?

Title: Mar 4 ­ 9:23 AM (17 of 20)

Page 18: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Is there an association between race and belief?Would this appear in a similar sample?

If the sample is perfectly independent, we have reason to believe the population is independent.

If the sample percentages are close but not equal, the sample is "somewhat dependent". We have some reason to suspect that the variables are dependent, but we can't be sure.

If the percentages are drastically different, we have strong reason to suspect that the variables are dependent.

Can we put a mathematical measure to our evidence?

Title: Mar 4 ­ 9:24 AM (18 of 20)

Page 19: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Title: Mar 4 ­ 9:24 AM (19 of 20)

Page 20: MATH 117 Applied Statistics is due Tuesday, March 6; and assignment …webspace.ship.edu/trcook/117/lesson/12-03-05.pdf ·  · 2012-03-04ME = 1.645*.056139 = .0923 e. ... (5 of 20)

Title: Mar 4 ­ 9:25 AM (20 of 20)